Oscillatory behaviour of a higher-order dynamic equation

Abstract

In this paper we are concerned with the oscillation of solutions of a certain more general higher-order nonlinear neutral-type functional dynamic equation with oscillating coefficients. We obtain some sufficient criteria for oscillatory behaviour of its solutions.

MSC:34N05.

1 Introduction

The calculus on time scales has been introduced in order to unify the theories of continuous and discrete processes and in order to extend those theories to a more general class of the so-called dynamic equations. In recent years there has been much research activity concerning the oscillation and non-oscillation of solutions of neutral dynamic equations on time scales.

In this paper we consider the higher-order nonlinear dynamic equation

$${[y(t)+P(t)y(\tau (t))]}^{{\mathrm{\Delta }}^n}+\sum _{i=1}^m{Q}_i(t)f_i\left(y({\varphi }_i(t))\right)=0,$$

(1.1)

where $n\ge 2$, $P(t),Q_i(t)\in C_{rd}{[t_0,\mathrm{\infty })}_{\mathbb{T}}$ for $i=1,2,\dots ,m$; $P(t)$ is an oscillating function ($P(t):\mathbb{T}\to \mathbb{R}$), $Q_i(t)$ are positive real-valued functions for $i=1,2,\dots ,m$; ${\varphi }_i(t)\in C_{rd}{[t_0,\mathrm{\infty })}_{\mathbb{T}}$, ${\varphi }_i^{\mathrm{\Delta }}(t)>0$, the variable delays $\tau ,{\varphi }_i:{[t_0,\mathrm{\infty })}_{\mathbb{T}}\to \mathbb{T}$ with $\tau (t),{\varphi }_i(t)<t$ for all $t\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, ${\varphi }_i(t)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$ for $i=1,2,\dots ,m$; $\tau (t)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$; $f_i(u)\in C(\mathbb{R},\mathbb{R})$ are nondecreasing functions, $u{f}_i(u)>0$ for $u\ne 0$ and $i=1,2,\dots ,m$.

The purpose of the paper is to study oscillatory behaviour of solutions of equation (1.1). For the sake of convenience, the function $z(t)$ is defined by

$$z(t)=y(t)+P(t)y(\tau (t)).$$

(1.2)

2 Basic definitions and some auxiliary lemmas

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers ℝ. For $t\in \mathbb{T}$, we define the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by

$$\sigma (t):=inf\{s\in \mathbb{T}:s>t\}$$

while the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ is defined by

$$\rho (t):=sup\{s\in \mathbb{T}:s<t\}.$$

If $\sigma (t)>t$, we say that t is right-scattered, while if $\rho (t)<t$, we say that t is left-scattered. Also, if $\sigma (t)=t$, then t is called right-dense, and if $\rho (t)=t$, then t is called left-dense. The graininess function $\mu :\mathbb{T}\to [0,\mathrm{\infty })$ is defined by

$$\mu (t):=\sigma (t)-t.$$

We introduce the set ${\mathbb{T}}^{\kappa }$ which is derived from the time scale $\mathbb{T}$ as follows. If $\mathbb{T}$ has left-scattered maximum m, then ${\mathbb{T}}^{\kappa }=\mathbb{T}-\{m\}$, otherwise ${\mathbb{T}}^{\kappa }=\mathbb{T}$.

Definition 1 [1]

The function $f:\mathbb{T}\to \mathbb{R}$ is called rd-continuous provided it is continuous at right-dense points in $\mathbb{T}$ and its left-sided limits exist (finite) at left-dense points in $\mathbb{T}$.

Theorem 1 [1]

Assume that $\nu :\mathbb{T}\to \mathbb{R}$ is strictly increasing and $\tilde{\mathbb{T}}:=\nu (\mathbb{T})$ is a time scale. Let $w:\tilde{\mathbb{T}}\to \mathbb{R}$. If ${\nu }^{\mathrm{\Delta }}(t)$ and $w^{\tilde{\mathrm{\Delta }}}(\nu (t))$ exist for $t\in {\mathbb{T}}^{\kappa }$, then

$${(w\circ \nu )}^{\mathrm{\Delta }}=(w^{\tilde{\mathrm{\Delta }}}\circ \nu ){\nu }^{\mathrm{\Delta }},$$

where we denote the derivative on $\tilde{\mathbb{T}}$ by $\tilde{\mathrm{\Delta }}$.

Definition 2 [1]

Let $f:\mathbb{T}\to \mathbb{R}$ be a function. If there exists a function $F:\mathbb{T}\to \mathbb{R}$ such that $F^{\mathrm{\Delta }}(t)=f(t)$ for all $t\in {\mathbb{T}}^{\kappa }$, then F is said to be an antiderivative of f. We define the Cauchy integral by

$${\int }_a^{b}f(\tau )\mathrm{\Delta }(\tau )=F(b)-F(a)\text{for }a,b\in \mathbb{T}.$$

Theorem 2 [2]

Let u and v be continuous functions on $[a,b]$ that are Δ-differentiable on $[a,b)$. If $u^{\mathrm{\Delta }}$ and $v^{\mathrm{\Delta }}$ are integrable from a to b, then

$${\int }_a^b{u}^{\mathrm{\Delta }}(t)v(t)\mathrm{\Delta }(t)+{\int }_a^b{u}^{\sigma }(t)v^{\mathrm{\Delta }}(t)\mathrm{\Delta }(t)=u(b)v(b)-u(a)v(a).$$

Let $\tilde{\mathbb{T}}=\mathbb{T}\cup \{sup\mathbb{T}\}\cup \{inf\mathbb{T}\}$. If $\mathrm{\infty }\in \tilde{\mathbb{T}}$, we call ∞ left-dense, and −∞ is called right-dense provided $-\mathrm{\infty }\in \tilde{\mathbb{T}}$. For any left-dense $t_0\in \tilde{\mathbb{T}}$ and any $\epsilon >0$, the set

$$L_{\epsilon }(t_0)=\{t\in \mathbb{T}:0<t_0-t<\epsilon \}$$

is nonempty, and so is $L_{\epsilon }(\mathrm{\infty })=\{t\in \mathbb{T}:t>\frac{1}{\epsilon }\}$ if $\mathrm{\infty }\in \tilde{\mathbb{T}}$.

Lemma 1 [3]

Let $n\in \mathbb{N}$ and f be n-times differentiable on $\mathbb{T}$. Assume $\mathrm{\infty }\in \tilde{\mathbb{T}}$. Suppose there exists $\epsilon >0$ such that

$$f(t)>0,sgn(f^{{\mathrm{\Delta }}^n}(t))\equiv s\in \{-1,+1\}\mathit{\text{for all}}t\in L_{\epsilon }(\mathrm{\infty })$$

and $f^{{\mathrm{\Delta }}^n}(t)\ne 0$ on $L_{\delta }(\mathrm{\infty })$ for any $\delta >0$. Then there exists $p\in [0,n]\cap {\mathbb{N}}_0$ such that $n+p$ is even for $s=1$ and odd for $s=-1$ with

$$\{\begin{array}{l}{(-1)}^{p+j}{f}^{{\mathrm{\Delta }}^j}(t)>0\mathit{\text{for all}}t\in L_{\epsilon }(\mathrm{\infty }),j\in [p,n-1]\cap {\mathbb{N}}_0,\\ f^{{\mathrm{\Delta }}^j}(t)>0\mathit{\text{for all}}t\in L_{{\delta }_j}(\mathrm{\infty })(\mathit{\text{with}}{\delta }_j\in (0,\epsilon )),j\in [1,p-1]\cap {\mathbb{N}}_0.\end{array}$$

Lemma 2 [3]

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^n}$, $t\in \mathbb{T}$, and $\alpha \in {\mathbb{T}}^{{\kappa }^n}$. Then with the functions $h_k$ defined as $h_n(t,s)={(-1)}^n{g}_n(s,t)$,

$$h_0(r,s)\equiv 1\mathit{\text{and}}{h}_{k+1}(r,s)={\int }_s^r{h}_k(\tau ,s)\mathrm{\Delta }s\mathit{\text{for}}k\in {\mathbb{N}}_0,$$

we have

$$f(t)=\sum _{k=0}^{n-1}{h}_k(t,\alpha )f^{{\mathrm{\Delta }}^k}(\alpha )+{\int }_{\alpha }^{{\rho }^{n-1}(t)}{h}_{n-1}(t,\sigma (\tau ))f^{{\mathrm{\Delta }}^n}(\tau )\mathrm{\Delta }\tau .$$

Lemma 3 [3]

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^n}$ and $m\in \mathbb{N}$ with $m<n$. Then we have, for all $\alpha \in {\mathbb{T}}^{{\kappa }^{n-1+m}}$ and $t\in {\mathbb{T}}^{{\kappa }^m}$,

$$f^{{\mathrm{\Delta }}^m}(t)=\sum _{k=0}^{n-m-1}{h}_k(t,\alpha )f^{{\mathrm{\Delta }}^{k+m}}(\alpha )+{\int }_{\alpha }^{{\rho }^{n-m-1}(t)}{h}_{n-m-1}(t,\sigma (\tau ))f^{{\mathrm{\Delta }}^n}(\tau )\mathrm{\Delta }\tau .$$

Lemma 4 [3]

Suppose f is n-times differentiable and $g_k$, $0\le k\le n-1$, are differentiable at $t\in {\mathbb{T}}^{{\kappa }^n}$ with

$$g_{k+1}^{\mathrm{\Delta }}(t)=g_k(\sigma (t))\mathit{\text{for all}}0\le k\le n-2.$$

Then we have

$${\left[\sum _{k=0}^{n-1}{(-1)}^k{f}^{{\mathrm{\Delta }}^k}{g}_k\right]}^{\mathrm{\Delta }}=f{g}_0^{\mathrm{\Delta }}+{(-1)}^{n-1}{f}^{{\mathrm{\Delta }}^n}{g}_{n-1}^{\sigma }.$$

3 Main results

Lemma 5 Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^n}$. If $f^{\mathrm{\Delta }}>0$, then for every λ, $0<\lambda <1$, we have

$$f(t)\ge \lambda {(-1)}^{n-1}{g}_{n-1}(\sigma \left(T^{\ast }\right),t)f^{{\mathrm{\Delta }}^{n-1}}(t).$$

(3.1)

Proof Let p, $0\le p\le n-1$, be the integer assigned to the function f as in Lemma 1. Because of $f^{\mathrm{\Delta }}>0$, we always have $p>0$. Furthermore, let $T^{\ast }\ge T$ be assigned to f by Lemma 1. Then, by using the Taylor formula on time scales, for every ${\rho }^{n-1}(t)\ge T^{\ast }$, we obtain

$$f(t)\ge {\int }_{T^{\ast }}^{{\rho }^{n-1}(t)}{(-1)}^{n-1}{g}_{n-1}(\sigma (\tau ),t)f^{{\mathrm{\Delta }}^n}(\tau )\mathrm{\Delta }\tau .$$

(3.2)

By using Theorem 2 and (3.2), we have

$$f(t)\ge {(-1)}^{n-1}{g}_{n-1}(\sigma (t),t)f^{{\mathrm{\Delta }}^{n-1}}(t)-{\int }_{T^{\ast }}^{{\rho }^{n-1}(t)}{(-1)}^{n-1}{g}_{n-1}(\sigma (\tau ),t)f^{{\mathrm{\Delta }}^{n-1}}(\tau )\mathrm{\Delta }\tau .$$

Since f is n-times differentiable on ${\mathbb{T}}^{{\kappa }^n}$ and $m\in \mathbb{N}$ with $m<n$, we have with n and f substituted by $n-m$ and $f^{{\mathrm{\Delta }}^m}$, respectively

$$f^{{\mathrm{\Delta }}^m}(t)\ge {\int }_{T^{\ast }}^{{\rho }^{n-m-1}(t)}{(-1)}^{n-m-1}{g}_{n-m-1}(\sigma (\tau ),t)f^{{\mathrm{\Delta }}^n}(\tau )\mathrm{\Delta }\tau .$$

Also, for every ${\rho }^{n-1}(t)$, s with ${\rho }^{n-1}(t)\ge T^{\ast }$ and $T^{\ast }\le s\le t$, we have

$$f^{{\mathrm{\Delta }}^m}(s)\ge {(-1)}^{n-m-1}{g}_{n-m-1}(\sigma \left(T^{\ast }\right),t)f^{{\mathrm{\Delta }}^n}(t).$$

This is obvious for $m=n-1$ and, when $m<n-1$, it can be derived by applying the Taylor formula. Thus, for all $t\ge T^{\ast }$, we get

$$f(t)\ge {(-1)}^{n-1}{g}_{n-1}(\sigma \left(T^{\ast }\right),t)f^{{\mathrm{\Delta }}^{n-1}}(t)$$

and therefore the proof of the lemma can be immediately completed. □

The result of Lemma 5 is an extension of studies in [4] and [5]. In order that the reader sees how the results in [4] (1.8.14) and [5] (Lemma 2) follow from (3.1), it is at this point only necessary to know that in the case $\mathbb{T}=\mathbb{Z}$, we have $\rho (t)=t-1$, $\sigma (t)=t+1$ and

$$g_{n-1}(\sigma \left(T^{\ast }\right),t)=\frac{{(t-T^{\ast }-1)}^{(n-1)}}{(n-1)!},$$

then we get the inequality in [4]

$$u(t)\ge \frac{1}{(n-1)!}{(n-n_1)}^{(n-1)}{\mathrm{\Delta }}^{n-1}u\left(2^{n-m-1}n\right);$$

and in the case $\mathbb{T}=\mathbb{R}$, we have $\rho (t)=\sigma (t)=t$ and

$$g_{n-1}(\sigma \left(T^{\ast }\right),t)=\frac{{(t-T^{\ast })}^{(n-1)}}{(n-1)!},$$

then we get the inequality in [5]

$$u(t)\ge \frac{\vartheta }{(n-1)!}{(t)}^{n-1}{u}^{n-1}(t).$$

For the cases $\mathbb{T}=\mathbb{Z}$ and $\mathbb{T}=\mathbb{R}$, some sufficient criterias for oscillatory behaviour of the solutions of the equation (1.1) were obtained by Bolat and Akın in [6] and [7], respectively. Furthermore, there might be other time scales that we cannot appreciate at this time due to our current lack of ‘real-world’ examples.

Theorem 3 Assume that n is odd and

(C1) ${lim}_{t\to \mathrm{\infty }}P(t)=0$,

(C2) ${\int }_{t_0}^{\mathrm{\infty }}{s}^{n-1}{\sum }_{i=1}^m{Q}_i(s)\mathrm{\Delta }s=\mathrm{\infty }$.

Then every bounded solution of equation (1.1) is either oscillatory or tends to zero as $t\to \mathrm{\infty }$.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y(t)$. Without loss of generality, assume that $y(t)$ is eventually positive (the proof is similar when $y(t)$ is eventually negative). That is, $y(t)>0$, $y(\tau (t))>0$ and $y({\varphi }_i(t))>0$ for $t\ge t_1\ge t_0$ and $i=1,2,\dots ,m$. Assume further that $y(t)$ does not tend to zero as $t\to \mathrm{\infty }$. By (1.1), (1.2), we have for $t\ge t_1$

$$z^{{\mathrm{\Delta }}^n}(t)=-\sum _{i=1}^m{Q}_i(t)f_i\left(y({\varphi }_i(t))\right)<0.$$

(3.3)

It follows that $z^{{\mathrm{\Delta }}^j}(t)$ ($j\in [0,n-1]\cap {\mathbb{N}}_0$) is strictly monotone and eventually of constant sign. Since $P(t)$ is an oscillatory function, there exists a $t_2\ge t_1$ such that if $t\ge t_2$, then $z(t)>0$. Since $y(t)$ is bounded, by virtue of (C1) and (1.2), there is a $t_3\ge t_2$ such that $z(t)$ is also bounded for $t\ge t_3$. Because n is odd and $z(t)$ is bounded, by Lemma 1, when $p=0$ (otherwise $z(t)$ is not bounded), there exists $t_4\ge t_3$ such that for $t\ge t_4$ we have ${(-1)}^j{z}^{{\mathrm{\Delta }}^j}(t)>0$, $j\in [0,n-1]\cap {\mathbb{N}}_0$.

In particular, since $z^{\mathrm{\Delta }}(t)<0$ for $t\ge t_4$, $z(t)$ is decreasing. Since $z(t)$ is bounded, we write ${lim}_{t\to \mathrm{\infty }}z(t)=L$ ($-\mathrm{\infty }<L<\mathrm{\infty }$). Assume that $0\le L<\mathrm{\infty }$. Let $L>0$. Then there exists a constant $c>0$ and a $t_5\ge t_4$ such that $z(t)>c>0$ for $t\ge t_5$. Since $y(t)$ is bounded, ${lim}_{t\to \mathrm{\infty }}P(t)y(\tau (t))=0$ by (C1). Therefore, there exists a constant $c_1>0$ and a $t_6\ge t_5$ such that $y(t)=z(t)-P(t)y(\tau (t))>c_1>0$ for $t\ge t_6$. So that we can find a $t_7$ with $t_7\ge t_6$ such that $y({\varphi }_i(t))>c_1>0$ for $t\ge t_7$. From (3.3) we have

$$z^{{\mathrm{\Delta }}^n}(t)=-\sum _{i=1}^m{Q}_i(t)f_i(c_1)<0$$

(3.4)

for $t\ge t_7$. If we multiply (3.4) by $t^{n-1}$ and integrate it from $t_7$ to t, we obtain

$$F(t)-F(t_7)\le -f(c_1){\int }_{t_7}^t\sum _{i=1}^m{Q}_i(s)s^{n-1}\mathrm{\Delta }s,$$

(3.5)

where

$$F(t)=\sum _{i=1}^{n-1}{(-1)}^{i+1}{\left(t^{n-1}\right)}^{{\mathrm{\Delta }}^i}{z}^{{\mathrm{\Delta }}^{n-i}}({\sigma }^i(t))$$

and

$${\sigma }^i(t)=\sigma ({\sigma }^{i-1}(t)).$$

Since ${(-1)}^k{z}^{{\mathrm{\Delta }}^k}(t)>0$ for $k=0,1,2,\dots ,n-1$ and $t\ge t_4$, we have $F(t)>0$ for $t\ge t_7$. From (3.5) we have

$$-F(t_7)\le -f(c_1){\int }_{t_7}^t\sum _{i=1}^m{Q}_i(s)s^{n-1}\mathrm{\Delta }s.$$

By (C2) we obtain

$$-F(t_7)\le -f(c_1){\int }_{t_7}^t\sum _{i=1}^m{Q}_i(s)s^{n-1}\mathrm{\Delta }s=-\mathrm{\infty }$$

as $t\to \mathrm{\infty }$. This is a contradiction. So, $L>0$ is impossible. Therefore, $L=0$ is the only possible case. That is, ${lim}_{t\to \mathrm{\infty }}z(t)=0$. Since $y(t)$ is bounded, by (C1) we obtain

$$\underset{t\to \mathrm{\infty }}{lim}y(t)=\underset{t\to \mathrm{\infty }}{lim}z(t)-\underset{t\to \mathrm{\infty }}{lim}P(t)y(t)=0$$

from (1.2).

Now let us consider the case of $y(t)<0$ for $t\ge t_1$. By (1.1) and (1.2),

$$z^{{\mathrm{\Delta }}^n}(t)=-\sum _{i=1}^m{Q}_i(t)f_i\left(y({\varphi }_i(t))\right)>0$$

for $t\ge t_1$. That is, $z^{{\mathrm{\Delta }}^n}>0$. It follows that $z^{{\mathrm{\Delta }}^j}(t)$ ($j\in [0,n-1]\cap {\mathbb{N}}_0$) is strictly monotone and eventually of constant sign. Since $P(t)$ is an oscillatory function, there exists a $t_2\ge t_1$ such that if $t\ge t_2$, then $z(t)<0$. Since $y(t)$ is bounded, by (C1) and (1.2), there is a $t_3\ge t_2$ such that $z(t)$ is also bounded for $t\ge t_3$. Assume that $x(t)=-z(t)$. Then $x^{{\mathrm{\Delta }}^n}(t)=-z^{{\mathrm{\Delta }}^n}(t)$. Therefore, $x(t)>0$ and $x^{{\mathrm{\Delta }}^n}(t)<0$ for $t\ge t_3$. Hence, we observe that $x(t)$ is bounded. Since n is odd, by Lemma 1, there exists a $t_4\ge t_3$ and $p=0$ (otherwise $x(t)$ is not bounded) such that ${(-1)}^j{x}^{{\mathrm{\Delta }}^j}(t)>0$, $j\in [0,n-1]\cap {\mathbb{N}}_0$ and $t\ge t_4$. That is, ${(-1)}^j{z}^{{\mathrm{\Delta }}^j}(t)<0$, $j\in [0,n-1]\cap {\mathbb{N}}_0$ and $t\ge t_4$. In particular, for $t\ge t_4$ we have $z^{\mathrm{\Delta }}(t)>0$. Therefore, $z(t)$ is increasing. So, we can assume that ${lim}_{t\to \mathrm{\infty }}z(t)=L$ ($-\mathrm{\infty }<L\le 0$). As in the proof of $y(t)>0$, we may prove that $L=0$. As for the rest, it is similar to the case of $y(t)>0$. That is, ${lim}_{t\to \mathrm{\infty }}y(t)=0$. This contradicts our assumption. Hence the proof is completed. □

Theorem 4 Assume that n is even and (C 1) holds. If the following condition is satisfied:

(C3) There is a function $\phi (t)$ such that $\phi (t)\in C_{rd}^1{[t_0,\mathrm{\infty })}_{\mathbb{T}}$. Moreover,

$$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{t_0}^t\phi (s)\sum _{i=1}^m{Q}_i(s)\mathrm{\Delta }s=\mathrm{\infty }$$

and

$$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{t_{10}}^t\frac{{[{\phi }^{\mathrm{\Delta }}(s)]}^2}{\phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))}\mathrm{\Delta }s<\mathrm{\infty }$$

for $\phi (t)$ and $i=1,2,\dots ,m$. Then every bounded solution of equation (1.1) is oscillatory.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y(t)$. Without loss of generality, assume that $y(t)$ is eventually positive (the proof is similar when $y(t)$ is eventually negative). That is, $y(t)>0$, $y(\tau (t))>0$ and $y({\varphi }_i(t))>0$ for $t\ge t_1\ge t_0$. By (1.1) and (1.2), we have (3.3) for $t\ge t_1$. Then $z^{{\mathrm{\Delta }}^n}(t)<0$. It follows that $z^{{\mathrm{\Delta }}^j}(t)$ ($j\in [0,n-1]\cap {\mathbb{N}}_0$) is strictly monotone and eventually of constant sign. Since $P(t)$ is an oscillatory function, there exists a $t_2\ge t_1$ such that for $t\ge t_2$, we have $z(t)>0$. Since $y(t)$ is bounded, by (C1) and (1.2), there is a $t_3\ge t_2$, such that $z(t)$ is also bounded for $t\ge t_3$. Because n is even, by Lemma 1 when $p=1$ (otherwise $z(t)$ is not bounded), there exists $t_4\ge t_3$ such that for $t\ge t_4$ we have

$${(-1)}^{j+1}{z}^{{\mathrm{\Delta }}^j}(t)>0,j\in [0,n-1]\cap {\mathbb{N}}_0.$$

(3.6)

In particular, since $z^{\mathrm{\Delta }}(t)>0$ for $t\ge t_4$, $z(t)$ is increasing. Since $y(t)$ is bounded,

$$\underset{t\to \mathrm{\infty }}{lim}P(t)y(\tau (t))=0$$

by (C1). Let $\delta >1$; i.e., there exists a $t_5\ge t_4$ such that by (1.2)

$$y(t)=z(t)-P(t)y(\tau (t))>\frac{1}{\delta }z(t)>0$$

for $t\ge t_5$. We may find a $t_6\ge t_5$ such that for $t\ge t_6$ and $i=1,2,\dots ,m$,

$$y({\varphi }_i(t))>\frac{1}{\delta }z({\varphi }_i(t))>0.$$

(3.7)

From (3.3), (3.7) and the properties of f, we have

$$\begin{array}{rl}{z}^{{\mathrm{\Delta }}^n}(t)& \le -\sum _{i=1}^m{Q}_i(t)f_i\left(\frac{1}{\delta }z({\varphi }_i(t))\right)\\ =-\sum _{i=1}^m{Q}_i(t)\frac{f_i(\frac{1}{\delta }z({\varphi }_i(t)))}{z({\varphi }_i(t))}z({\varphi }_i(t))\end{array}$$

(3.8)

for $t\ge t_6$. Since $z(t)>0$ is bounded and increasing, ${lim}_{t\to \mathrm{\infty }}z(t)=L$ ($0<L<\mathrm{\infty }$). By the continuity of f, we have

$$\underset{t\to \mathrm{\infty }}{lim}\frac{f_i(\frac{1}{\delta }z({\varphi }_i(t)))}{z({\varphi }_i(t))}=\frac{f_i(\frac{L}{\delta })}{L}>0.$$

Then there is a $t_7\ge t_6$ such that for $t\ge t_7$, $i=1,2,\dots ,m$, we have

$$\frac{f_i(\frac{1}{\delta }z({\varphi }_i(t)))}{z({\varphi }_i(t))}\ge \frac{f_i(\frac{L}{\delta })}{2L}=\alpha >0.$$

(3.9)

By (3.8), (3.9),

$$z^{{\mathrm{\Delta }}^n}(t)\le -\alpha \sum _{i=1}^m{Q}_i(t)z({\varphi }_i(t))\text{for }t\ge t_7.$$

(3.10)

Set

$$w(t)=\frac{z^{{\mathrm{\Delta }}^{n-1}}(t)}{z(\frac{1}{\delta }{\varphi }_i(t))}.$$

(3.11)

We know from (3.6) that there is a $t_8\ge t_7$ such that for a sufficiently large $t\ge t_8$, $w(t)>0$. Therefore, Δ-derivating (3.11) we obtain

$$\begin{array}{rl}{w}^{\mathrm{\Delta }}(t)& =\frac{z^{{\mathrm{\Delta }}^n}(t)}{z({\delta }^{-1}{\varphi }_i(t))}+z^{{\mathrm{\Delta }}^{n-1}}(\sigma (t)){\left(\frac{1}{z({\delta }^{-1}{\varphi }_i(t))}\right)}^{\mathrm{\Delta }}\\ =\frac{z^{{\mathrm{\Delta }}^n}(t)}{z({\delta }^{-1}{\varphi }_i(t))}-\frac{{\delta }^{-1}{\varphi }_i^{\mathrm{\Delta }}(t)z^{{\mathrm{\Delta }}^{n-1}}(\sigma (t))z^{\mathrm{\Delta }}({\delta }^{-1}{\varphi }_i(t))}{z({\delta }^{-1}{\varphi }_i(t))z({\delta }^{-1}{\varphi }_i(\sigma (t)))}\end{array}$$

(3.12)

$$\begin{array}{r}\le \frac{z^{{\mathrm{\Delta }}^n}(t)}{z({\delta }^{-1}{\varphi }_i(t))}-\frac{{\delta }^{-1}{\varphi }_i^{\mathrm{\Delta }}(t)z^{{\mathrm{\Delta }}^{n-1}}(\sigma (t))z^{\mathrm{\Delta }}({\delta }^{-1}{\varphi }_i(t))}{z^2({\delta }^{-1}{\varphi }_i(\sigma (t)))}\\ =\frac{z^{{\mathrm{\Delta }}^n}(t)}{z({\delta }^{-1}{\varphi }_i(t))}-{\delta }^{-1}{w}^{\sigma }(t)\frac{{\varphi }_i^{\mathrm{\Delta }}(t)z^{\mathrm{\Delta }}({\delta }^{-1}{\varphi }_i(t))}{z({\delta }^{-1}{\varphi }_i(\sigma (t)))}.\end{array}$$

(3.13)

We know from (3.6) that there is a $t\ge t_9$ such that $z^{\mathrm{\Delta }}(t)>0$ and $z^{{\mathrm{\Delta }}^{n-1}}(t)>0$ for an even n. Since $z(t)>0$ is increasing $z({\delta }^{-1}{\varphi }_i(\sigma (t)))\ge z({\delta }^{-1}{\varphi }_i(t))$ for $i=1,2,\dots ,m$. Therefore, by Lemma 5,

$$z({\delta }^{-1}{\varphi }_i(t))\ge \lambda {(-1)}^{n-1}{g}_{n-1}(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}({\varphi }_i(t)).$$

(3.14)

Then by Δ-derivating (3.14) and using $g_{n-1}^{\mathrm{\Delta }}(\sigma (t),t)=g_{n-2}^{\sigma }(\sigma (t),t)$, we get

$$\begin{array}{rl}{\left[z({\delta }^{-1}{\varphi }_i(t))\right]}^{\mathrm{\Delta }}& \ge \lambda {(-1)}^{n-2}{g}_{n-1}^{\mathrm{\Delta }}(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}({\varphi }_i(t))\\ \ge \lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}({\varphi }_i(t))\end{array}$$

by Lemma 2

$$z^{\mathrm{\Delta }}({\delta }^{-1}{\varphi }_i(t)){\delta }^{-1}{\varphi }_i^{\mathrm{\Delta }}(t)\ge \lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}({\varphi }_i(t)).$$

Since ${\varphi }_i(t)\le t$, we obtain

$$z^{\mathrm{\Delta }}({\delta }^{-1}{\varphi }_i(t))\ge \frac{\delta \lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}(t)}{{\varphi }_i^{\mathrm{\Delta }}(t)}.$$

(3.15)

Hence by (3.10), (3.14) and (3.15), we have

$$\begin{array}{rl}{w}^{\mathrm{\Delta }}(t)\le & \frac{-\alpha {\sum }_{i=1}^m{Q}_i(t)z({\varphi }_i(t))}{z({\delta }^{-1}{\varphi }_i(t))}\\ -{\delta }^{-1}{w}^{\sigma }(t)\frac{\delta \lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t))z^{{\mathrm{\Delta }}^{n-1}}(t)}{{\varphi }_i^{\mathrm{\Delta }}(t)}\frac{{\varphi }_i^{\mathrm{\Delta }}(t)}{z({\delta }^{-1}{\varphi }_i(\sigma (t)))}\\ \le & \frac{-\alpha {\sum }_{i=1}^m{Q}_i(t)z({\varphi }_i(t))}{z({\delta }^{-1}{\varphi }_i(t))}\\ -{\delta }^{-1}{w}^{\sigma }(t)\frac{\delta \lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t)){\varphi }_i^{\mathrm{\Delta }}(t)}{{\varphi }_i^{\mathrm{\Delta }}(t)}\frac{z^{{\mathrm{\Delta }}^{n-1}}(\sigma (t))}{z({\delta }^{-1}{\varphi }_i(\sigma (t)))}\\ \le & -\alpha \sum _{i=1}^m{Q}_i(t)-\lambda {(-1)}^{n-2}{g}_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t)){(w^{\sigma }(t))}^2,\end{array}$$

and then

$$\alpha \sum _{i=1}^m{Q}_i(t)\le -w^{\mathrm{\Delta }}(t)-\lambda {(-1)}^{n-2}{w}^2(t)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(t)),{\varphi }_i(t))$$

(3.16)

for $t\ge t_{10}$. If we multiply (3.16) by $\phi (t)$ and integrate it from $t_{10}$ to t, we obtain by Theorem 2

$$\begin{array}{rl}\alpha {\int }_{t_{10}}^t\phi (s)\sum _{i=1}^m{Q}_i(s)\mathrm{\Delta }s\le & -{\int }_{t_{10}}^t\phi (s)w^{\mathrm{\Delta }}(s)\mathrm{\Delta }s\\ -{\int }_{t_{10}}^t\lambda {(-1)}^{n-2}\phi (s)w^2(s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))\mathrm{\Delta }s\\ \le & -[\phi (t)w(t)-\phi (t_{10})w(t_{10})-{\int }_{t_{10}}^t{\phi }^{\mathrm{\Delta }}(s)w^{\sigma }(t)\mathrm{\Delta }s]\\ -{\int }_{t_{10}}^t\lambda {(-1)}^{n-2}\phi (s)w^2(s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))\mathrm{\Delta }s\\ \le & \phi (t_{10})w(t_{10})+{\int }_{t_{10}}^t{\phi }^{\mathrm{\Delta }}(s)w^{\sigma }(t)\mathrm{\Delta }s\\ -\lambda {\int }_{t_{10}}^t\phi (s)w^2(s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))\mathrm{\Delta }s\\ \le & \phi (t_{10})w(t_{10})-\lambda {\int }_{t_{10}}^t\phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))\\ \times {[w(s)-\frac{{\phi }^{\mathrm{\Delta }}(s)}{2\lambda \phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))}]}^2\mathrm{\Delta }s\\ +{\int }_{t_{10}}^t\frac{{[{\phi }^{\mathrm{\Delta }}(s)]}^2}{4\lambda \phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))}\mathrm{\Delta }s\\ \le & \phi (t_{10})w(t_{10})+{\int }_{t_{10}}^t\frac{{[{\phi }^{\mathrm{\Delta }}(s)]}^2}{4\lambda \phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))}\mathrm{\Delta }s.\end{array}$$

Therefore by (C3),

$$\begin{array}{rl}\mathrm{\infty }& =\alpha \underset{t\to \mathrm{\infty }}{lim}sup{\int }_{t_{10}}^t\phi (s)\sum _{i=1}^m{Q}_i(s)\mathrm{\Delta }s\\ \le \phi (t_{10})w(t_{10})+\frac{1}{4\lambda }\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{t_{10}}^t\frac{{[{\phi }^{\mathrm{\Delta }}(s)]}^2}{\phi (s)g_{n-2}^{\sigma }(\sigma ({\varphi }_i(s)),{\varphi }_i(s))}\mathrm{\Delta }s\\ <\mathrm{\infty }.\end{array}$$

This is a contradiction.

Now let us consider the case of $y(t)<0$ for $t\ge t_1$. By (1.1) and (1.2), we have

$$z^{{\mathrm{\Delta }}^n}(t)=-\sum _{i=1}^m{Q}_i(t)f_i\left(y({\varphi }_i(t))\right)>0$$

for $t\ge t_1$. That is, $z^{{\mathrm{\Delta }}^n}>0$. It follows that $z^{{\mathrm{\Delta }}^j}(t)$ ($j\in [0,n-1]\cap {\mathbb{N}}_0$) is strictly monotone and eventually of constant sign. Since $P(t)$ is an oscillatory function, there exists a $t_2\ge t_1$ such that $z(t)<0$ for $t\ge t_2$. Since $y(t)$ is bounded, by (C1) and (1.2), there is a $t_3\ge t_2$ such that $z(t)$ is also bounded for $t\ge t_3$. Assume that $x(t)=-z(t)$. Then $x^{{\mathrm{\Delta }}^n}(t)=-z^{{\mathrm{\Delta }}^n}(t)$. Therefore, $x(t)>0$ and $x^{{\mathrm{\Delta }}^n}(t)<0$ for $t\ge t_3$. Hence, we observe that $x(t)$ is bounded. Since n is odd, by Lemma 1, there exists a $t_4\ge t_3$ and $p=1$ (otherwise $x(t)$ is not bounded) such that ${(-1)}^k{x}^{{\mathrm{\Delta }}^k}(t)>0$, $k\in [0,n-1]\cap {\mathbb{N}}_0$ and $t\ge t_4$. That is, ${(-1)}^k{z}^{{\mathrm{\Delta }}^k}(t)<0$, $k\in [0,n-1]\cap {\mathbb{N}}_0$ and $t\ge t_4$. In particular, for $t\ge t_4$ we have $z^{\mathrm{\Delta }}(t)>0$. Therefore, $z(t)$ is increasing. For the rest of the proof, we can proceed the proof similarly to the case of $y(t)>0$. Hence, the proof is completed. □